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Pani-rosa [81]
3 years ago
5

Can someone help me on this

Chemistry
1 answer:
xz_007 [3.2K]3 years ago
8 0
155,500



I did this to the best of my ability. I have a hard time comprehending things sometimes so I’m so so so sorry if it’s wrong
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How many molecules are in 100 g of C6H120,?*​
Grace [21]

Answer:

3.37 × 10²³ molecules

Explanation:

Given data:

Mass of C₆H₁₂O₆ = 100 g

Number of molecules = ?

Solution:

Number of moles of C₆H₁₂O₆:

Number of moles = mass/molar mass

Number of moles = 100 g/ 180.16 g/mol

Number of moles = 0.56 mol

Number of molecules:

1 mole contain 6.022 × 10²³ molecules

0.56 mol × 6.022 × 10²³ molecules /1 mol

3.37 × 10²³ molecules

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Ethylenediamine (en) forms an octahedral complex with Ni2+(aq) with the formula [Ni(en)3]2+. Ni2+(aq) + 3 en ⇌ [Ni(en)3]2+(aq) K
yawa3891 [41]

Answer:

3.125\times 10^{-19} mol/L is the concentration of Ni^{2+}(aq) in the solution.

Explanation:

Ni^{2+}(aq) + 3 en\rightleftharpoons [Ni(en)_3]^{2+}(aq)

Concentration of nickel ion = [Ni^{2+}]=x

Concentration of nickel complex= [[Ni(en)_3]^{2+}]=\frac{0.16 mol}{2 L}=0.08 mol/L

Concentration of ethylenediamine = [en]=\frac{0.80 mol}{2 L}=0.40 mol/L

The formation constant of the complex = K_f=4.0\times 10^{18}

The expression of formation constant is given as:

K_f=\frac{[[Ni(en)_3]^{2+}]}{[Ni^{2+}][en]^3}

4.0\times 10^{18}=\frac{0.08 mol/L}{x\times (0.40 mol/L)^3}

x=\frac{0.08 mol/L}{4.0\times 10^{18}\times (0.40 mol/L)^3}

x=3.125\times 10^{-19} mol/L

3.125\times 10^{-19} mol/L is the concentration of Ni^{2+}(aq) in the solution.

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